Math 181, Exam 1, Study Guide
Problem 1 Solution
1. Evaluate
!
2
xex dx.
Solution: We evaluate the integral using the u-substitution method. Let u = x2 . Then
1
du = x dx and we get:
du = 2x dx ⇒
2
!
!
2
x2
xe dx = ex · x dx
!
1
= eu · du
2
!
1
=
eu du
2
1
= eu + C
2
1 2
= ex + C
2
1
Math 181, Exam 1, Study Guide
Problem 2 Solution
2. Evaluate
!
1
16x2
+1
dx.
Solution: We evaluate the integral using the u-substitution method. Let u = 4x. Then
1
du = dx and we get:
du = 4 dx ⇒
4
!
!
1
1
dx =
dx
2
16x + 1
(4x)2 + 1
!
1
1
=
· du
2
u +1 4
!
1
1
=
du
4
u2 + 1
1
= arctan u + C
4
1
arctan(4x) + C
=
4
1
Math 181, Exam 1, Study Guide
Problem 3 Solution
3. Evaluate
!
√
t3
dt.
t4 + 9
Solution: We evaluate the integral using the u-substitution method. Let u = t4 + 9. Then
1
du = 4t3 dt ⇒
du = t3 dt and we get:
4
!
!
t3
1
√
√
· t3 dt
dt =
4+9
t4 + 9
t
!
1 1
√ · du
=
u 4
!
1
1
√ du
=
4
u
1" √ #
= 2 u +C
4
1√
u+C
=
2
1√ 4
t +9+C
=
2
1
Math 181, Exam 1, Study Guide
Problem 4 Solution
4. Evaluate
!
tan2 θ sec2 θ dθ.
Solution: We evaluate the integral using the u-substitution method. Let u = tan θ. Then
du = sec2 θ dθ and we get:
!
!
2
2
tan θ sec θ dθ = u2 du
1
= u3 + C
3
1
=
tan3 θ + C
3
1
Math 181, Exam 1, Study Guide
Problem 5 Solution
5. Evaluate
!
1
0
√
x 1 − x2 dx.
Solution: We evaluate the integral using the u-substitution method. Let u = 1 − x2 . Then
1
du = −2x dx ⇒ − du = x dx. The limits of integration becomes u = 1 − 02 = 1 and
2
u = 1 − 12 = 0. We get:
! 1 √
! 1√
x 1 − x2 dx =
1 − x2 (x dx)
0
0
#
"
! 0
√
1
=
u − du
2
1
! 0
1
=−
u1/2 du
2 1
!
1 1 1/2
=
u du
2 0
$
%1
1 2 3/2
u
=
2 3
0
$
%
$
%
1 2 3/2
1 2 3/2
=
(1)
−
(0)
2 3
2 3
=
1
3
1
Math 181, Exam 1, Study Guide
Problem 6 Solution
6. Evaluate
!
π
−π
√
cos x
dx.
4 + 3 sin x
Solution: We evaluate the integral using the u-substitution method. Let u = 4 + 3 sin x.
1
du = cos x dx. The limits of integration become u = 4 +
Then du = 3 cos x dx ⇒
3
3 sin(−π) = 4 and u = 4 + 3 sin π = 4. We get:
! π
! π
cos x
1
√
√
dx =
· cos x dx
4 + 3 sin x
4 + 3 sin x
−π
−π
! 4
1 1
√ · du
=
u 3
4
= 0
because the limits of integration are identical.
1
Math 181, Exam 1, Study Guide
Problem 7 Solution
7. Evaluate
!
1
0
√
4
dx.
1 − x2
Solution: The integrand is undefined at x = 1. Therefore, this is an improper integral. We
evaluate the integral by turning it into a limit calculation.
! 1
! b
4
4
√
√
dx = lim−
dx
2
b→1
1−x
1 − x2
0
0
"
#b
= lim− 4 arcsin x
b→1
0
= lim− [4 arcsin b − 4 arcsin 0]
b→1
= 4 arcsin 1 − 4 arcsin 0
$π %
− 4(0)
=4
2
= 2π
We evaluated the limit lim− 4 arcsin b by substituting b = 1 using the fact that f (b) =
b→1
4 arcsin b is left-continuous at b = 1.
1
Math 181, Exam 1, Study Guide
Problem 8 Solution
dy
for each of the following:
dx
! x√
(a) y =
1 + t2 dt
8. Find
0
(b) y =
!
(c) y =
!
√
0
x
" #
sin t2 dt
tan x
0
1
dt (hint: when you simplify, you will get a constant)
1 + t2
Solution: In all parts, we use the Fundamental Theorem of Calculus Part II:
! g(x)
d
f (t) dt = f (g(x))g ′(x)
dx a
(a) The derivative is:
! x√
dy
d
=
1 + t2 dt
dx
dx 0
$
= 1 + (x)2 · (x)′
√
=
1 + x2
(b) The derivative is:
dy
d
=
dx
dx
!
√
x
" #
sin t2 dt
0
%"√
#2 & "√ #′
= sin
x
x
·
1
= sin x · √
2 x
(c) The derivative is:
dy
d
=
dx
dx
!
0
tan x
1
dt
1 + t2
1
· (tan x)′
2
1 + (tan x)
1
· sec2 x
=
1 + tan2 x
1
=
· sec2 x
2
sec x
= 1
=
1
Math 181, Exam 1, Study Guide
Problem 9 Solution
9. Find the area of the region enclosed by the curves y = x2 and y = −x2 + 4x.
Solution:
y
4
3
2
1
1
0
2
x
The formula we will use to compute the area of the region is:
! b
Area =
(top − bottom) dx
a
where the limits of integration are the x-coordinates of the points of intersection of the two
curves. These are found by setting the y’s equal to each other and solving for x.
y=y
x2 = −x2 + 4x
2x2 − 4x = 0
2x(x − 2) = 0
x = 0, x = 2
From the graph we see that the top curve is y = −x2 + 4x and the bottom curve is y = x2 .
1
Therefore, the area is:
Area =
=
!
!
b
(top − bottom) dx
a
2
!0 2
"#
$ # $%
−x2 + 4x − x2 dx
$
4x − 2x2 dx
0
'2
&
2 3
2
= 2x − x
3
0
2
= 2(2)2 − (2)3
3
8
=
3
=
#
2
Math 181, Exam 1, Study Guide
Problem 10 Solution
10. Find the area of the region enclosed by the curves x + y 2 = 0 and x + 3y 2 = 2.
Solution:
y
1
0
-1
1
2
x
-1
The formula we will use to compute the area of the region is:
! d
Area =
(right − left) dx
c
where the limits of integration are the y-coordinates of the points of intersection of the two
curves. These are found by setting the x’s equal to each other and solving for y.
x=x
−y 2 = −3y 2 + 2
2y 2 − 2 = 0
2(y 2 − 1) = 0
2(y + 1)(y − 1) = 0
y = −1, y = 1
From the graph we see that the right curve is x = −3y 2 + 2 and the left curve is x = −y 2 .
1
Therefore, the area is:
!
d
1
=
!
=
!
Area =
(right − left) dx
c
−1
1
−1
"#
#
$ #
$%
−3y 2 + 2 − −y 2 dy
$
2 − 2y 2 dy
'1
&
2 3
= 2y − y
3
−1
' &
'
&
2 3
2
3
= 2(1) − (1) − 2(−1) − (−1)
3
3
=
8
3
2
Math 181, Exam 1, Study Guide
Problem 11 Solution
11. Find the volume of the following solid:
The base of the solid is the circle x2 + y 2 = 1. The cross-sections are isosceles right triangles
perpendicular to the y-axis.
Solution: The volume formula we will use is:
Z d
V =
A(y) dy
c
where A(y) is the cross sectional area of the solid as a function of y. We choose to integrate
with respect to y in this problem because the cross sections are perpendicular to the y-axis.
To determine the function A(y) we first note that the cross sections are isosceles right
triangles. The triangle may be oriented in one of two ways: (1) with the hypotenuse lying
inside the circle x2 + y 2 = 1 or (2) with one of the sides lying inside the circle.
Let’spassume (1) for the moment. In this case, the length of the hypotenuse is 2x where
x = 1 y 2 after solving the equation x2 + y 2 = 1 for x. The other two sides of the triangle
are equal since the triangle is isosceles. Letting the other sides be a, we use the Pythagorean
Theorem to find a.
a2 + a2
2a2
2a2
a2
= (2x)2
= 4x2
= 4(1 y 2 )
= 2(1 y 2 )
p
a = 2(1 y 2 )
1
The area of the triangle is:
1
A(y) = bh
2
1
= (a)(a)
2
1
= a2
2
⇤
1⇥
=
2 1 y2
2
= 1 y2
The cross sections start at c =
1 and end at d = 1. Therefore, the volume is:
Z 1
V =
1 y 2 dy

= y

= 1
=1
=
1
1
1 3
y
3
1 3
(1)
3
1
+1
3
1
1
3

( 1)
1
( 1)3
3
4
3
Now let’s assume (2),pthat one of the sides lies in the circle. In this case, the length of the
side is 2x where x = 1 y 2 . The other side of the triangle is also 2x since the triangle is
2
isosceles. The area of the triangle is then:
1
A(y) = bh
2
1
= (2x)(2x)
2
= 2x2
= 2(1 y 2 )
The cross sections start at c =
V =
1 and end at d = 1. Therefore, the volume is:
Z
1
2 1
1

=2 y

=2 1
✓
=2 1
=
y 2 dy
1
1 3
y
3
1

1 3
(1)
( 1)
3
◆
1
1
+1
3
3
8
3
3
1
( 1)3
3
Math 181, Exam 1, Study Guide
Problem 12 Solution
12. Find the volume of the solid generated by revolving the region bounded by y =
y = 2, and x = 0 about the:
√
x,
(a) x-axis
(b) y-axis
Solution:
(a) We find the volume of the solid generated by revolving around the x-axis using the
Washer Method. The variable of integration is x and the corresponding formula is:
! b
"
#
V =π
(top)2 − (bottom)2 dx
a
√
The top curve is y = 2 and the bottom curve is y = x. The lower limit of integration
is x = 0. The upper
limit is the x-coordinate of the point of intersection of the curves
√
y = 2 and y = x. To find this, we set the y’s equal to each other and solve for x.
y=y
√
x=2
x=4
The volume is then:
! 4$
%√ &2 '
2
V =π
(2) −
x
dx
0
! 4
=π
(4 − x) dx
0
)4
(
1 2
= π 4x − x
2
0
)
(
1 2
= π 4(4) − (4)
2
= 8π
1
y
2
1
0
1
2
3
4
x
(b) We find the volume of the solid generated by revolving around the y-axis using the
Shell Method. The variable of integration is x and the corresponding formula is:
! b
V = 2π
x (top − bottom) dx
a
The top and bottom curves are the same as those in part (a). So are the limits of
integration. The volume is then:
! 4
%
√ &
V = 2π
x 2 − x dx
0
! 4
%
&
= 2π
2x − x3/2 dx
0
)4
(
2 5/2
2
= 2π x − x
5
0
(
)
2
2
5/2
= 2π 4 − (4)
5
=
32π
5
2
Math 181, Exam 1, Study Guide
Problem 13 Solution
13. Find the volume of the solid generated by revolving the region bounded by the y-axis
2
and the curve x = , for 1 ≤ y ≤ 4, about the:
y
(a) x-axis, using the Method of Cylindrical Shells
(b) y-axis
Solution:
(a) We find the volume of the solid obtained by rotating about the x-axis using the Shell
Method. The variable of integration is y and the corresponding formula is:
! d
V = 2π
y (right − left) dy
c
The right curve is x =
2
and the left curve is x = 0 (the y-axis). The volume is then:
y
#
! 4 "
2
V = 2π
y
− 0 dy
y
1
! 4
= 2π
2 dy
1
$ %4
= 2π 2y
1
= 2π [2(4) − 2(1)]
= 12π
y
4
3
2
1
0
1
1
2
x
(b) We find the volume of the solid obtained by rotating about the y-axis using the Disk
Method. The variable of integration is y and the corresponding formula is:
! d
V =π
f (y)2 dy
c
2
where f (y) = . The volume is then:
y
4
V =π
!
4
=π
!
1
" #2
2
dy
y
4y −2 dy
& '4
1
= 4π −
y
&" 1# " #'
1
1
− −
= 4π −
4
1
1
= 3π
2
Math 181, Exam 1, Study Guide
Problem 14 Solution
14. In some chemical reactions, the rate at which the amount of a substance changes with
time is proportional to the amount present. Consider a substance whose amount obeys the
equation:
dy
= −0.6y
dt
where t is measured in hours. If there are 100 grams of the substance present when t = 0,
how many grams will be left after 1 hour?
Solution: The amount of the substance y(t) is given by the formula:
y(t) = y0 e−0.06t
where y0 = 100 grams is the initial amount of the substance. After 1 hour, the amount of
the substance is:
y(1) = 100e−0.06 grams
1
Math 181, Exam 1, Study Guide
Problem 15 Solution
dy
is proportional
15. Suppose the rate at which the number of people infected with a disease
dt
to the number of people currently infected y:
dy
= ky
dt
Suppose that, in the course of any given year, the number of people infected is reduced by
20%. If there are 10,000 infected people today, how many years will it take to reduce the
number to 1000?
Solution: The number of people infected is given by the function:
y(t) = y0 ekt
where y0 = 10, 000 is the initial number of people infected. To answer the question in the
problem, we need to find the value of k. Since the number of people infected is reduced by
20% in the course of any given year, the number of people infected after the first year is:
10, 000 − 0.20(10, 000) = 8, 000
This corresponds to the value y(1). Using the function above for y(t) we get:
y(1) = 10, 000ek(1)
8, 000 = 10, 000ek
8, 000
ek =
10, 000
4
ek =
5
4
k = ln
5
To find how many years it will take for the number of infected people to reduce to 1,000, we
set y(t) equal to 1,000 and solve for t.
4
y(t) = 10, 000e(ln 5 )t
4
1, 000 = 10, 000e(ln 5 )t
4
1, 000
= e(ln 5 )t
10, 000
4
1
= e(ln 5 )t
10 !
"
1
4
= ln
t
ln
10
5
t=
1
ln 10
ln 45
1
Math 181, Exam 1, Study Guide
Problem 16 Solution
16. Consider the definite integral:
!
0
4
"
#
x2 + x dx
(a) Compute the exact value of the integral.
(b) Estimate the value of the integral using the Trapezoidal Rule with N = 4.
(c) Estimate the value of the integral using Simpson’s Rule with N = 4.
(d) Which of the above two estimate is more accurate?
Solution:
(a) The exact value is:
!
0
4
$
%4
" 2
#
1 3 1 2
x + x dx = x + x
3
2
0
1 3 1 2
= (4) + (4)
3
2
88
=
3
(b) Using N = 4, the length of each subinterval of [0, 4] is:
∆x =
4−0
b−a
=
=1
N
4
The estimate T4 is:
∆x
[f (0) + 2f (1) + 2f (2) + 2f (3) + f (4)]
2
'
1& 2
(0 + 0) + 2(12 + 1) + 2(22 + 2) + 2(32 + 3) + (42 + 4)
=
2
1
= [0 + 4 + 12 + 24 + 20]
2
= 30
T4 =
1
(c) The estimate S4 is:
∆x
[f (0) + 4f (1) + 2f (2) + 4f (3) + f (4)]
3
'
1& 2
=
(0 + 0) + 4(12 + 1) + 2(22 + 2) + 4(32 + 3) + (42 + 4)
3
1
= [0 + 8 + 12 + 48 + 20]
3
88
=
3
S4 =
(d) Clearly, S4 is more accurate because it is the exact value of the integral.
2
Math 181, Exam 1, Study Guide
Problem 17 Solution
17. Consider the definite integral:
Estimate the value of the integral using:
!
2
0
dx
1 + x2
(a) the Trapezoidal Rule with N = 2
(b) the Midpoint method with N = 2
(c) Simpson’s Rule with N = 4
Solution:
(a) The length of each subinterval of [0, 2] is:
∆x =
b−a
2−0
=
=1
N
2
The estimate T2 is:
∆x
[f (0) + 2f (1) + f (2)]
2"
#
1
1
1
1
=
+2·
+
2 1 + 02
1 + 12 1 + 22
"
#
1
1
=
1+1+
2
5
T2 =
=
11
10
(b) The length of each subinterval of [0, 2] is ∆x = 1 just as in part (a). The estimate M2
is:
" $ %
$ %#
1
3
M2 = ∆x f
+f
2
2
)
&
1
1
=1·
' (2 +
' (2
1 + 21
1 + 23
4
4
= +
5 13
72
=
65
1
(c) We can use the following formula to find S4 :
1
2
S4 = M2 + T2
3
3
where M2 and T2 were found in parts (a) and (b). We get:
$ %
$ %
1 11
2 72
+
S4 =
3 65
3 10
48 11
=
+
65 30
431
=
390
2